Virtual Gate Test Series: Linear Algebra - Rank Of The Matrix

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edited

Plz see the picture below for the detailed explanation.

$A_{m\times n}$, So  $A^{T}_{n\times m}$

let $m=2,n=1$ and assume some values in the matrix.

$A_{2\times 1}=\begin{bmatrix} 1\\2 \end{bmatrix}$ and $A^{T}_{1\times2}=\begin{bmatrix} 1 &2 \end{bmatrix}$

Here rank is $r(A)=1$ [apply some rows operation in matrix and get the rank,and rank is the maximum number of independent rows/columns]
$A_{2\times 1}.A^{T}_{1\times 2}=\begin{bmatrix} 1\\2 \end{bmatrix}.\begin{bmatrix} 1 &2 \end{bmatrix}$

$A_{2\times 1}.A^{T}_{1\times 2}=\begin{bmatrix} 1 &2 \\2 &4 \end{bmatrix}$

Here rank is $r(A_{2\times1}A^{T}_{1\times2})=1$  [apply some rows operation in matrix and get the rank]

Given that rank of the $A_{m\times n}$is equal to $e$

So,rank of $A_{m\times n}.A^{T}_{n\times m}=e$

edited

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